Match the story to the shape: Before fitting anything, look at how the quantity behaves and pick the family that matches.
| Family | When it fits |
|---|---|
| Straight line y = mx + c | A steady, constant change (same rise per step). |
| Quadratic y = ax² + bx + c | One turning point (a projectile's height, a profit that peaks). |
| Cubic | Two turning points (an up-down-up trend). |
| Exponential y = k·aˣ (or k·erx) | A quantity that multiplies by a fixed factor each step (growth a > 1, decay 0 < a < 1). |
| Power y = a·xⁿ | Scaling laws (area vs length, allometry in biology). |
| Logistic y = L / (1 + C·e−kx) | Growth that starts fast then levels off at a ceiling L (populations, market saturation). |
| Sinusoidal y = a·sin(b(x − c)) + d | Anything that repeats (tides, temperature over a year, daylight hours). |
| Piecewise | One rule on one interval, a different rule on another (tax bands, bulk-discount pricing). |
The two tell-tale questions: Does it level off at a ceiling? → logistic (S-shape), not plain exponential.
Does it repeat? → sinusoidal.
These are the two that students most often miss — an exponential keeps climbing forever, but a real population can't, so a 'fast then flattens' graph is logistic.
IB-style question — pick the model
A conservation team releases beavers into a wetland. The first few yearly counts rise quickly, but the team knows the wetland can only support a finite number.
Which function family best models the beaver population over time, and why?
Step by step
- Fast early rise → growth, so NOT linear and NOT a decay.
- A finite ceiling (carrying capacity) → it must flatten, ruling out plain exponential.
- An S-shaped curve that flattens at a maximum L is the logistic model.
Final answer
A logistic model y = L/(1 + C·e−kt), because the population grows fast at first then levels off at the wetland's carrying capacity L. (A plain exponential would be wrong — it never stops growing.)
Free preview
This is the free notes preview
You're reading the free notes. Aimnova Pro unlocks the full study experience — and you can try it with your first topic free to keep:
- FlashcardsLock in vocabulary and key terms with spaced repetition.
- Practice questionsAnswer exam-style questions and get instant AI marking.
- Mock exams & past-paper vaultSit full mocks and see exactly how examiners award marks.
- Personalised study planA daily plan built around your exam date and weak areas.
Fit, then translate every parameter into words: Once you've chosen a family, find the parameters — by hand from given conditions, or with your GDC's regression. The marks come from saying what each one means in the scenario.
What each parameter means
- In y = k·ert: k is the starting value (t = 0) and r is the continuous growth/decay rate.
- In the logistic y = L/(1 + C·e−kt): L is the long-run ceiling (carrying capacity), and k controls how fast it gets there.
- In a·sin(b(t − c)) + d: a = amplitude (half the swing), d = midline (average level), period = 2π/b, c = horizontal shift.
Animated graph
Watch the graph build step by step in study mode.
IB-style question — fit an exponential and interpret
A colony of bacteria is modelled by N(t) = k·ert, where N is the number of cells (in thousands) and t is the time in hours. At t = 0 there are 3 thousand cells; after 5 hours there are 24 thousand.
Find k and r, and state what r tells you about the colony.
Step by step
- At t = 0: N = k·e⁰ = k, so k is the starting value.
- At t = 5: 24 = 3·e5r, so e5r = 8.
- Take ln of both sides and solve.
- r > 0 means continuous growth at about 41.6% per hour.
Final answer
k = 3 (the colony starts at 3 thousand cells) and r ≈ 0.416. Since r > 0, the colony is growing, at a continuous rate of about 41.6% per hour.
IB-style question — read the logistic ceiling
A fish population in a new reservoir is modelled by P(t) = 8000/(1 + 15·e−0.5t), where t is in years.
State the long-term population the model predicts, and find the initial population.
Step by step
- Long-term: as t → ∞, e−0.5t → 0, so the denominator → 1.
- Initial: put t = 0, so e⁰ = 1.
Final answer
Long-term (carrying capacity) L = 8000 fish; the reservoir is stocked with 500 fish initially.